WLOG $a+b=1$. Let $ab=m$. Then $1^2-m=3c^2$ and $1^3-2\times1\cdot m=4d^3$. So $\begin{cases}1-m=3c^2,\\1-2m=4d^3.\end{cases}$ Meaning that $6c^2=4d^3+1$. Notice that $$ab=m\le\left(\frac{a+b}2\right)^2=\frac14,$$ so $4d^3\ge1-\frac14\times2$, so $d\ge\frac12$. Our inequality becomes $1+d\le3c$. Squaring and substituting implies that proving $$12d^3-2d^2-4d+1\ge0$$ should suffice. Factorize this into $(2d-1)(6d^2+2d-1)$ and we see that it is true.

youthdoo wrote: WLOG $a+b=1$. Let $ab=m$. Then $1^2-m=3c^2$ and $1^3-2\times1\cdot m=4d^3$. So $\begin{cases}1-m=3c^2,\\1-2m=4d^3.\end{cases}$ Meaning that $6c^2=4d^3+1$. Notice that $$ab=m\le\left(\frac{a+b}2\right)^2=\frac14,$$ so $4d^3\ge1-\frac14\times2$, so $d\ge\frac12$. Our inequality becomes $1+d\le3c$. Squaring and substituting implies that proving $$12d^3-2d^2-4d+1\ge0$$ should suffice. Factorize this into $(2d-1)(6d^2+2d-1)$ and we see that it is true. Why could you assume a+b=1? Is there a name for this technique?

Because the variables are to the same power in each equation or inequality.

parmenides51 wrote: Given real positive numbers $a, b, c$ and $d$, for which the equalities $a^2 + ab + b^2 = 3c^2$ and $a^3 + a^2b + ab^2 + b^3 = 4d^3$ are fulfilled. Prove that $$a + b + d \le 3c.$$ Reducing $c$ and $d$, the inequality to prove can be written as $$a+b+\sqrt[3]{\frac{(a+b)(a^2+b^2)}{4}} \leq \sqrt{3(a^2+ab+b^2)}$$ Using Holder's inequality and $\frac{3}{4}(a+b)^2 \leq a^2+ab+b^2$: $$\begin{aligned} 2 \cdot \frac{a+b}{2}+\sqrt[3]{\frac{(a+b)(a^2+b^2)}{4}} &\leq \sqrt[3]{9\left[\frac{(a+b)^3}{8}+\frac{(a+b)^3}{8}+\frac{(a+b)(a^2+b^2)}{4}\right]}\\ &=\sqrt[3]{\frac{9}{4}(a+b)\left[(a+b)^2+a^2+b^2\right]}\\ &= \sqrt[3]{\frac{9}{2}(a+b)(a^2+ab+b^2)}\\ & \leq \sqrt{3(a^2+ab+b^2)}\\ \end{aligned}$$